A104975 -id:A104975 - cp's OEIS frontend

A104976 Row sums of A104975.

1, 1, 0, 0, 1, 1, -1, -1, 2, 2, -2, -2, 4, 4, -6, -6, 9, 9, -13, -13, 21, 21, -31, -31, 47, 47, -71, -71, 109, 109, -165, -165, 250, 250, -380, -380, 578, 578, -876, -876, 1330, 1330, -2020, -2020, 3068, 3068, -4656, -4656, 7070, 7070, -10736, -10736, 16300, 16300, -24746, -24746, 37574, 37574, -57050, -57050

Offset: 0

Paul Barry, Mar 30 2005

G. C. Greubel, Table of n, a(n) for n = 0..1000

Cf. A104975, A104977.

t[n_, k_]:= t[n, k]= If[k==n, 1, ((1+(-1)^(n-k))/2)*Sum[Binomial[k, j]*t[(n-k)/2, j], {j,(n-k)/2}] ];
S[n_]:= Sum[(-1)^j*t[n, j], {j,0,n}]; (* S = A104977 *)
a[n_]:= a[n]= Sum[If[EvenQ[n-k], S[(n-k)/2], 0], {k,0,n}];
Table[a[n], {n, 0, 65}] (* G. C. Greubel, Jun 08 2021 *)

@CachedFunction
def t(n,k): return 1 if (k==n) else ((1+(-1)^(n-k))/2)*sum( binomial(k, j)*t((n-k)/2, j) for j in (1..(n-k)//2) )
def S(n): return sum( (-1)^j*t(n, j) for j in (0..n) ) # S = A104977
def T(n,k): return S((n-k)/2) if (mod(n-k, 2)==0) else 0 # T = A104975
def a(n): return sum( T(n,k) for k in (0..n) )
[a(n) for n in (0..65)] # G. C. Greubel, Jun 08 2021

a(n) = Sum_{k=0..n} A104975(n, k).

G.f.: x^2/((1-x)*(Sum_{k>=1} x^(2^k))).

A104977 Defining sequence for an inverse Fredholm-Rueppel triangle.

Original entry on oeis.org

1, -1, 1, -2, 3, -4, 6, -10, 15, -22, 34, -52, 78, -118, 180, -274, 415, -630, 958, -1454, 2206, -3350, 5088, -7724, 11726, -17806, 27036, -41046, 62320, -94624, 143668, -218130, 331191, -502854, 763486, -1159206, 1760038, -2672286, 4057356, -6160326, 9353294, -14201206, 21561836

Offset: 0

Paul Barry, Mar 30 2005

A104975 is the sequence array for this sequence.

abs(a(n)) is the numbers of compositions of n into parts of the form 2^k-1, see example. [Joerg Arndt, Jan 06 2013]

			G.f. = 1 - x + x^2 - 2*x^3 + 3*x^4 - 4*x^5 + 6*x^6 - 10*x^7 + 15*x^8 - 22*x^9 + 34*x^10 + ...
From _Joerg Arndt_, Jan 06 2013: (Start)
There are abs(a(8)) = 15 compositions of 8 into parts 2^k-1:
[ 1]  [ 1 1 1 1 1 1 1 1 ]
[ 2]  [ 1 1 1 1 1 3 ]
[ 3]  [ 1 1 1 1 3 1 ]
[ 4]  [ 1 1 1 3 1 1 ]
[ 5]  [ 1 1 3 1 1 1 ]
[ 6]  [ 1 1 3 3 ]
[ 7]  [ 1 3 1 1 1 1 ]
[ 8]  [ 1 3 1 3 ]
[ 9]  [ 1 3 3 1 ]
[10]  [ 1 7 ]
[11]  [ 3 1 1 1 1 1 ]
[12]  [ 3 1 1 3 ]
[13]  [ 3 1 3 1 ]
[14]  [ 3 3 1 1 ]
[15]  [ 7 1 ]
(End)

Robert Israel, Table of n, a(n) for n = 0..5509
Johann Cigler, Relation between a continued fraction and partitions, mathoverflow question 292231, Feb 05 2018.
Johann Cigler, A curious class of Hankel determinants, arXiv:1803.05164 [math.CO], 2018.

Cf. A090678, A209229.

N:= 100: # to get a(0)..a(N)
S:= series(x/add(x^(2^k),k=0..ilog2(N+1)),x,N+2):
[seq](coeff(S, x, j), j = 0 .. N); # Robert Israel, Feb 07 2018

T[n_, n_] = 1; T[n_, m_] := T[n, m] =(1 + (-1)^(n-m))/2 Sum[Binomial[m, k]* T[(n-m)/2, k], {k, 1, (n-m)/2}];
a[n_] := Sum[T[n, m] (-1)^m, {m, 0, n}];
Array[a, 50, 0] (* Jean-François Alcover, Jul 21 2018, after Vladimir Kruchinin *)

T(n,m):=if n=m then 1 else (1+(-1)^(n-m))/2*sum(binomial(m,k)*T((n-m)/2,k),k,1,(n-m)/2); makelist(sum(T(n,m)*(-1)^(m),m,0,n),n,0,20); /* Vladimir Kruchinin, Mar 18 2015 */

N = 66;  q = 'q + O('q^N);  L = 2+ceil(log(N)/log(2));
gf = q / sum(n=0, L, q^(2^n) );
/* gf = 1 / (1 + sum(n=1, L, q^(2^n-1) ) ); */  /* same */
v = Vec(gf)
/* Joerg Arndt, Jan 06 2013 */

G.f.: x / (sum_{k>=0} x^(2^k)). (corrected by Joerg Arndt, Jan 06 2013)
G.f.: 1 / (1 + sum(n>=1, x^(2^n-1) ) ), replace the '+' by '-' to obtain the g.f. for compositions into parts 2^k-1. [Joerg Arndt, Jan 06 2013]
G.f.: 1 - x / (1 + x / (1 + x / (1 - x / (1 + x / (1 - x / ...))))) = 1 + b(1) * x / (1 + b(2) * x / (1 + b(3) * x / ...)) where b(n) = (-1)^ A090678(n+1) [Conjecture]. - Michael Somos, Jan 03 2013
Convolution inverse is A209229 with 0 preprended. - Michael Somos, Jan 03 2013
a(n) = sum(m=0..n, T(n,m)*(-1)^(m)), where T(n,m)=(1+(-1)^(n-m))/2 *sum(k=1..(n-m)/2, binomial(m,k)*T((n-m)/2,k)), T(n,n)=1. - Vladimir Kruchinin, Mar 18 2015

Added more terms, Joerg Arndt, Jan 06 2013

A104974 A Fredholm-Rueppel triangle.

Original entry on oeis.org

1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1

Offset: 0

Paul Barry, Mar 30 2005

Sequence matrix for A036987(n+1).
Riordan array ( (Sum_{k>=0} x^(2^k)/x^2) - 1/x, x).
Diagonal sums are A070939(n+1), with interpolated zeros.
Inverse is A104975.

			Triangle begins as:
  1;
  0, 1;
  1, 0, 1;
  0, 1, 0, 1;
  0, 0, 1, 0, 1;
  0, 0, 0, 1, 0, 1;
  1, 0, 0, 0, 1, 0, 1;
  0, 1, 0, 0, 0, 1, 0, 1;
  0, 0, 1, 0, 0, 0, 1, 0, 1;
  0, 0, 0, 1, 0, 0, 0, 1, 0, 1;
  0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1;
  0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1;
  0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1;
  0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

Cf. A000523 (row sums), A036987, A070939, A104975.

[(Catalan(n-k+1) mod 2): k in [0..n], n in [0..15]]; // G. C. Greubel, Jun 08 2021

A104974 := proc(n,k)
    modp(A000108(n+1-k),2);
end proc:
seq(seq( A104974(n,k), k=0..n), n=0..15); # R. J. Mathar, Apr 21 2021

Table[Mod[CatalanNumber[n-k+1], 2], {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jun 08 2021 *)

flatten([[mod(catalan_number(n-k+1), 2) for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jun 08 2021

T(n, k) = A000108(n+1-k) mod 2. [Corrected by R. J. Mathar, Apr 21 2021]

Sum_{k=0..n} T(n, k) = A000523(n+1).

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A104976 Row sums of A104975.

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A104977 Defining sequence for an inverse Fredholm-Rueppel triangle.

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A104974 A Fredholm-Rueppel triangle.

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